If one zero of the polynomial (a^{2} + 9) x^{2} + 13x + 6a is the reciprocal of the other, find the value of a.

We have,

(a + 9) x^{2} – 13x + 6a = 0

Here, we have:

A = (a^{2} + 9), B = 13 and C = 6a

Let us assume be the zeros of the given polynomial

∴ Product =

=

1 =

a^{2} + 9 = 6a

a^{2} – 6a + 9 = 0

a^{2} – 2 × a × 3 + 3^{2} = 0

(a – 3)^{2} = 0

a – 3 = 0

a = 3

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